Cho x, y, z là ba số thực dương, thỏa mãn \(xy+yz+...

- Hướng dẫn giải

Phương pháp giải:

Sử dụng BĐT Cô-si.

Giải chi tiết:

Đặt \(T=\frac{xy}{{{z}^{3}}\left( 1+x \right)\left( 1+y \right)}+\frac{yz}{{{x}^{3}}\left( 1+y \right)\left( 1+z \right)}+\frac{zx}{{{y}^{3}}\left( 1+z \right)\left( 1+x \right)}\)

Từ giả thiết \(xy+yz+zx=xyz\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Áp dụng BĐT Cô-si ta có:

\(\begin{align}  & \frac{xy}{{{z}^{3}}\left( 1+x \right)\left( 1+y \right)}+\frac{1+x}{64x}+\frac{1+y}{64y}\ge 3\sqrt[3]{\frac{xy}{{{z}^{3}}\left( 1+x \right)\left( 1+y \right)}.\frac{1+x}{64x}.\frac{1+y}{64y}}=\frac{3}{16z} \\  & \frac{yz}{{{x}^{3}}\left( 1+y \right)\left( 1+z \right)}+\frac{1+y}{64y}+\frac{1+z}{64z}\ge 3\sqrt[3]{\frac{yz}{{{x}^{3}}\left( 1+y \right)\left( 1+z \right)}.\frac{1+y}{64y}.\frac{1+z}{64z}}=\frac{3}{16x} \\  & \frac{zx}{{{y}^{3}}\left( 1+z \right)\left( 1+z \right)}+\frac{1+z}{64z}+\frac{1+x}{64x}\ge 3\sqrt[3]{\frac{zx}{{{y}^{3}}\left( 1+z \right)\left( 1+x \right)}.\frac{1+z}{64z}.\frac{1+x}{64x}}=\frac{3}{16y} \\  & \Rightarrow A+\frac{2}{64}\left( \frac{1+x}{x}+\frac{1+y}{y}+\frac{1+z}{z} \right)\ge \frac{3}{16}\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \\  & \Leftrightarrow A+\frac{1}{32}\left( 3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\ge \frac{3}{16}\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \\  & \Leftrightarrow A+\frac{1}{8}\ge \frac{3}{16} \\  & \Leftrightarrow A\ge \frac{1}{16} \\ \end{align}\)

Dấu bằng xảy ra \(\Leftrightarrow \left\{ \begin{align}  & \frac{xy}{{{z}^{3}}\left( 1+x \right)\left( 1+y \right)}=\frac{1+x}{64x}=\frac{1+y}{64y} \\  & \frac{yz}{{{x}^{3}}\left( 1+y \right)\left( 1+z \right)}=\frac{1+y}{64y}=\frac{1+z}{64z} \\  & \frac{zx}{{{y}^{3}}\left( 1+z \right)\left( 1+z \right)}=\frac{1+z}{64z}=\frac{1+x}{64x} \\  & \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \\ \end{align} \right.\Leftrightarrow x=y=z=3\)